if x1=8sin theta and x2= 6 cos theta then prove that maximum value of x1 + x2 is 10.

we have

x1 = 8sinθ

x2 = 6cosθ

so,

x = x1 + x2 = 8sinθ + 6cosθ

to find maximum value of 'x' we differentiate it wrt 'θ' and set it to zero. so,

dx/dv = [d/dθ](8sinθ + 6cosθ) = 0

or

8cosθ - 6sinθ = 0

or

4cosθ = 3sinθ

thus,

tanθ = 4/3

or the value of x1 + x2 is maximum when

θ = 53.129

so,

(max) x = (max) x1 + x2 = 8sin(53.129) + 6cos(53.129)

or

(max) x1 + x2 = 8x0.799 + 6x0.6 

(max) x1 + x2 ~ 6.4 + 3.6

thus, maximum values of x1 + x2 will be = 10